//给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。 
//
// 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。 
//
// 
//
// 示例 1： 
//
// 
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"ABCCED"
//输出：true
//
//
// 示例 2：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"SEE"
//输出：true
//
//
// 示例 3：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"ABCB"
//输出：false
//
//
//
//
// 提示：
//
//
// m == board.length
// n = board[i].length
// 1 <= m, n <= 6
// 1 <= word.length <= 15
// board 和 word 仅由大小写英文字母组成
//
//
//
//
// 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
// Related Topics 数组 回溯 矩阵 👍 1139 👎 0

package leetcode.editor.cn;

public class WordSearch{
    public static void main(String[] args) {

//        Solution solution = new WordSearch().new Solution();
    char[][] board = {{'A','B','C','E'},{'S','F','C','S'},{'A','D','E','E'}};
    String word = "ABCCED";
    System.out.println(new WordSearch().new Solution().exist(board, word));
    }
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public boolean exist(char[][] board, String word) {
        if (board == null || board[0] == null) return false;
//        if (word.length() < 1 || word.length() > 15) return false;
        char[] words = word.toCharArray();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
               if (dfs(board, words, i, j, 0)) return true;
            }
        }
        return false;
    }

    public boolean dfs(char[][] board, char[] words, int i, int j, int k) {
        if (i >= board.length || i < 0 || j >= board[0].length || j < 0 || words[k] != board[i][j]) return false;
        if (k == words.length - 1) return true;
        char temp = board[i][j];
        board[i][j] = '\0';
        boolean res = dfs(board, words, i + 1, j, k + 1) || dfs(board, words, i - 1, j, k + 1)
                || dfs(board, words, i, j + 1, k + 1) || dfs(board, words, i, j - 1, k + 1);
        board[i][j] = temp;
        return res;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

}